\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 145, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/145\hfil Oscillation for dynamic equations]
{Oscillation for forced second-order nonlinear dynamic equations
on time scales}

\author[M. Huang, W. Feng \hfil EJDE-2006/145\hfilneg]
{Mugen Huang, Weizhen Feng}

\address{School of Mathematical Sciences, South China Normal
University, Guangzhou 510631,  China}
 \email[M. Huang]{huangmugen@sohu.com}
\email[W. Feng]{wsy@scnu.edu.cn}

\date{}
\thanks{Submitted August 18, 2006. Published November 26, 2006.}
\subjclass[2000]{34K11, 39A10, 39A99, 34C10, 39A11}
\keywords{Forced oscillation; dynamic equations; time scales}

\begin{abstract}
 By means of Riccati transformation techniques, we present
 oscillation criteria for forced second-order nonlinear
 dynamic equations on time scales. These results are based
 on the information on a sequence of subintervals of
 $[a, \infty)$ only, rather than on the whole half-line.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}


\section{Introduction}

The theory of time scales, which has recently received a lot of
attention, was introduced by Hilger \cite{Hilger} in his Ph.D. Thesis
in 1988 in order to unify continuous and discrete analysis.
A time scale $\mathbb{T}$, is an arbitrary nonempty closed subset of
the reals. Many authors have expanded on various aspects of this
new theory; see the survey paper by Agarwal et al. \cite{Agarwal}
and the book by Bohner and Peterson \cite{Bohner1} which summarizes
and organizes much of the time scale calculus. For the notion used
below we refer to the next section that provides some basic facts on
time scale extracted from \cite{Bohner1}.

There are many interesting time scales and they give rise to plenty
of applications, the cases when the time scale is equal to reals or
the integers represent the classical theories of differential and of
difference equations. Another useful time scale is
$\mathbb{P}_{a,b}=\cup^{\infty}_{n=0}[n(a+b), n(a+b)+a]$ which
is widely used to study population in biological communities,
electric circuit and so on.

In recent years, there has been much research activity concerning
the oscillation and nonoscillation of solution of various equations
on time scales, and we refer the reader to papers
\cite{Bep,Bohner2,Erbe,Sahiner,Saker2,Saker1} and references cited
therein.

 Bohner and Saker\cite{Bohner2} considered the perturbed
nonlinear dynamic equation
\begin{equation}
(\alpha(t)(x^{\Delta})^{\gamma})^{\Delta}+F(t, x^{\sigma})=G(t,
x^{\sigma}, x^{\Delta}), \quad t\in [a, b].
\label{e1.1}
\end{equation}
Assuming that $\frac{F(t, u)}{f(u)}\geq q(t), \frac{G(t, u,
v)}{f(u)}\leq p(t)$, they change \eqref{e1.1} into the  inequality
\begin{equation}
(\alpha(t)(x^{\Delta})^{\gamma})^{\Delta}+(q(t)-p(t))f(x^{\sigma})\leq0.
\label{e1.2}
\end{equation}
Then using Riccati transformation techniques, they obtain
sufficient conditions for the solution to be oscillatory, or to
converge to zero.

 Saker \cite{Saker2} considered the second-order forced nonlinear
dynamic equation
$$
(a(t)x^{\Delta})^{\Delta}+p(t)f(x^{\sigma})=r(t), \quad t\in [t_0,
\infty),
\label{e1.3}
$$
assuming that $\int^{\infty}_{t_0}|r(s)|\Delta s<\infty$; that
is, the forcing terms are ``small'' enough for all large $t\in
\mathbb{T}$. Some additional assumptions have to be imposed on the
unknown solutions. He obtained sufficient condition on the forcing
terms directly, for solution to be oscillatory or to
converge to zero.

Following this trend, to develop the qualitative theory of dynamic
equations on time scales, in this paper, we consider the following
second-order forced nonlinear dynamic equation
\begin{equation}
x^{\Delta \Delta}(t)+p(t)f(x^{\sigma}(t))=e(t),
\label{e1.4}
\end{equation}
on the time scale interval $[a, \infty)=\{t\in \mathbb{T}, t\geq a\}$,
where $x^{\sigma}(t)=x(\sigma (t))$, $e, p\in C_{\rm rd}(\mathbb{T},
\mathbb{R})$.

In this paper, we apply Riccati transformation technique to obtain
some oscillation criteria for \eqref{e1.4}. Our results do not require that
$p(t)$ and $e(t)$ be of definite sign and are based on the
information only on a sequence of subintervals of $[a, \infty)$
rather than the whole half-line. Our results in this paper improve
the results given in \cite{Bohner2,Saker2}.

By a solution of \eqref{e1.4}, we mean a nontrivial real-valued function
$x$ satisfying \eqref{e1.4} for $t\geq a$. A solution $x$ of \eqref{e1.4} is
called oscillatory if it is neither eventually positive nor
eventually negative; otherwise it is called nonoscillatory.
Equation \eqref{e1.4}
is called oscillatory if all solutions are oscillatory. Our
attention is restricted to those solution $x$ of \eqref{e1.4} which
exist on half line $[t_{x}, \infty)$ with $\sup\{|x(t)|: t\geq
t_0\}\neq 0$ for any $t_0\geq t_{x}$.


\section{Preliminaries}

Let $\mathbb{T}$ be a time scale, we define the forward and backward
jump operators by
$$
\sigma(t)=\inf\{s\in \mathbb{T}: {s>t}\}, \quad
\rho(t)=\sup\{s\in \mathbb{T}: s<t\},
$$
where $\inf \emptyset=\sup \mathbb{T}$, $\sup \emptyset=\inf \mathbb{T}$, and
$\emptyset$ denotes the empty set. A nonmaximal element $t\in \mathbb{T}$
is called right-dense if $\sigma(t)=t$ and right-scattered if
$\sigma(t)>t$. A nonminimal element $t\in \mathbb{T}$ is said to be
left-dense if $\rho(t)=t$ and left-scattered if $\rho(t)<t$. The
graininess $\mu$ of the time scale $\mathbb{T}$ is defined by
$\mu(t)=\sigma(t)-t$.

A mapping $f: \mathbb{T}\to \mathbb{X}$ is said to be differentiable
at $t\in \mathbb{T}$, if there exists $b\in \mathbb{X}$ such that
for every $\varepsilon>0$, there exists a neighborhood $U$ of
$t$ satisfying $|[f(\sigma(t))-f(s)]-b[\sigma(t)-s]|\leq \varepsilon
|\sigma(t)-s|$, for all $s\in U$. We say that $f$ is delta
differentiable (or in short: differentiable) on $\mathbb{T}$
provided $f^{\Delta}(t)$ exist for all $t\in \mathbb{T}$.

A function $f: \mathbb{T}\to \mathbb{R}$ is called rd-continuous
provided it is continuous at right-dense points in $\mathbb{T}$ and
its left-sided limits exist (finite) at left-dense points in
$\mathbb{T}$. The set of rd-continuous functions
$f: \mathbb{T}\to \mathbb{R}$ will be denoted by
$C_{\rm rd}(\mathbb{T}, \mathbb{R})$.

The derivative and forward jump operator $\sigma$ are related by the
formula
\begin{equation}
f(\sigma(t))=f(t)+\mu(t)f^{\Delta}(t).
\label{e2.1}
\end{equation}
Let $f$ be a differentiable function on $[a,b]$.
If $f^{\Delta}>0$, $f^{\Delta}<0$, $f^{\Delta}\geq 0$,
$f^{\Delta}\leq 0$ for all $t\in [a, b)$;
then $f$ is increasing, decreasing, nondecreasing, nonincreasing on $[a, b]$,
respectively.

We  use the following product  and quotient rules for derivative of
two differentiable functions $f$ and $g$
\begin{gather}
(fg)^{\Delta}=f^{\Delta}g+f^{\sigma}g^{\Delta}=fg^{\Delta}+f^{\Delta}g^{\sigma},
\label{e2.2}
\\
\big(\frac{f}{g}\big)^{\Delta}=\frac{f^{\Delta}g-fg^{\Delta}}{gg^{\sigma}},
\label{e2.3}
\end{gather}
where $f^{\sigma}=f\circ \sigma$ and $gg^{\sigma}\neq 0$.

The integration by parts formula reads
\begin{equation}
\int^{b}_{a}f^{\Delta}(t)g(t)\Delta t
=f(t)g(t)|^{b}_{a}-\int^{b}_{a}f^{\sigma}(t)g^{\Delta}(t)\Delta t,
\label{e2.4}
\end{equation}

Chain Rule: Assume $g: \mathbb{T}\to \mathbb{R}$ is
$\Delta-$differentiable on $\mathbb{T}$ and
$f: \mathbb{R}\to \mathbb{R}$ is continuously differentiable.
Then $f\circ g: \mathbb{T}\to \mathbb{R}$ is $\Delta$-differentiable
and satisfies
\begin{equation}
(f\circ g)^{\Delta}(t)=\Big\{\int^{1}_{0}f'(g(t)
+h\mu (t)g^{\Delta}(t))dh\Big\}g^{\Delta}(t). \label{e2.5}
\end{equation}
 From \eqref{e2.5}, we obtain (see \cite{Saker1}).
\begin{equation}
(x^{\gamma})^{\Delta}(t)
=\gamma\int^{1}_{0}[hx^{\sigma}(t)+(1-h)x(t)]^{\gamma-1}dhx^{\Delta}(t).
\label{e2.6}
\end{equation}
In order to prove our main results, we need the following auxiliary
result.

\begin{lemma} \label{lem1}
If $A$ and $B$ are nonnegative, then
\begin{equation}
A^{\lambda}-\lambda AB^{\lambda-1}+(\lambda-1)B^{\lambda}\geq 0, \quad
\lambda >1,
\label{e2.7}
\end{equation}
and the equality holds if and only if $A=B$.
\end{lemma}


\section{Main results}

Our interest is to establish oscillation criteria for \eqref{e1.4} that
do not assume that $p(t)$ and $e(t)$ being of definite sign. In this
section, we give some new oscillation criteria. Since
we are interested in oscillatory behavior, we suppose that the time
scale $\mathbb{T}$ under consideration is not bounded above, i.e. it
is a time scale interval of the form $[a, \infty)$.
Let
$$
D(a_i, b_i)=\big\{u\in C^{1}_{\rm rd}[a_i, b_i]: u(t)\not\equiv 0,
u(a_i)=u(b_i)=0\big\}, \quad i=1,2.
$$

\begin{theorem}\label{th1}
Let $f(x)/x\geq k>0$ for $x\neq 0$. Assume that for any
$T\geq a$, there exist constants $a_1, b_1, a_2, b_2\in \mathbb{T}$
such that $T\leq a_1<b_1$, $T\leq a_2<b_2$, and
\begin{equation}
\begin{gathered}
p(t)\geq 0, \quad t\in [a_1, b_1]\cup [a_2, b_2],\\
e(t)\leq 0, \quad t\in [a_1, b_1]; \\
e(t)\geq 0, \quad t\in[a_2,b_2].
\end{gathered} \label{e3.1}
\end{equation}
If there exists $u\in D(a_i, b_i)$ such that
\begin{equation}
\int^{b_i}_{a_i}\Big\{kp(t)u^2(\sigma(t))
-\frac{\mu(t)+t-a_i}{4(t-a_i)}
\big[\frac{u(t)+u(\sigma(t))}{u(\sigma(t))}u^{\Delta}(t)\big]^2\Big\}\Delta
t\geq 0, \label{e3.2}
\end{equation}
for $i=1, 2$, then \eqref{e1.4} is oscillatory.
\end{theorem}

\begin{proof}
 Suppose, to the contrary, that $x$ is a nonoscillatory
solution of \eqref{e1.4}, which is eventually positive. Say $x(t)>0$,
$x^{\sigma}(t)>0$ for $t\geq t_0\geq a$. Denote
$w(t)=-\frac{x^{\Delta}(t)}{x(t)}$ for $t\geq t_0$. It follows from
\eqref{e1.4} that $w(t)$ satisfies the dynamic equation
\begin{equation}
\begin{aligned}
w^{\Delta}(t)&=-\frac{x^{\Delta
\Delta}(t)}{x^{\sigma}(t)}+\frac{x(t)}{x^{\sigma}(t)}
\big[\frac{x^{\Delta}(t)}{x(t)}\big]^2\\
&=\frac{1}{1+\mu(t)\frac{x^{\Delta}(t)}{x(t)}}w^2(t)
+p(t)\frac{f(x^{\sigma}(t))}{x^{\sigma}(t)}-\frac{e(t)}{x^{\sigma}(t)}.
\end{aligned}\label{e3.3}
\end{equation}
By assumption, we can choose $a_1, b_1\in \mathbb{T}$ such that
$b_1>a_1\geq t_0$ and $p(t)\geq 0, e(t)\leq 0, t\in [a_1, b_1]$.
 From \eqref{e1.4}, we get
$x^{\Delta\Delta}(t)=e(t)-p(t)f(x^{\sigma}(t))\leq 0$ for
$t\in [a_1, b_1]$. Therefore, we have that for $t\in [a_1, b_1]$
$$
x(t)\geq x(t)-x(a_1)=\int^{t}_{a_1}x^{\Delta}(s)\Delta s
\geq x^{\Delta}(t)(t-a_1);
$$
i.e.,
\begin{equation}
\frac{x^{\Delta}(t)}{x(t)}\leq \frac{1}{t-a_1},  \quad t\in (a_1, b_1].
\label{e3.4}
\end{equation}
Using the above inequality and $\frac{f(x)}{x}\geq k$, \eqref{e3.3} yields
\begin{equation}
w^{\Delta}(t)\geq \frac{t-a_1}{\mu(t)+t-a_1}w^2(t)+kp(t).
\label{e3.5}
\end{equation}
Let $u(t)\in D(a_1, b_1)$ be as in the hypothesis. Multiply
both sides of \eqref{e3.5} by $u^2(\sigma(t))$ and integrate it from $a_1$
to $b_1$, we obtain
\begin{equation}
\int^{b_1}_{a_1}u^2(\sigma(t))w^{\Delta}(t)\Delta t \geq
\int^{b_1}_{a_1}\left[\lambda_1(t)w^2(t)u^2(\sigma(t))
+kp(t)u^2(\sigma(t))\right]\Delta t, \label{e3.6}
\end{equation}
where $\lambda_1(t)=\frac{t-a_1}{\mu(t)+t-a_1}$. Using the
integration by parts, \eqref{e2.4}, and $u(a_1)=u(b_1)=0$, we have
\begin{equation}
\begin{aligned}
0&=w(t)u^2(t)|^{b_1}_{a_1}\\
&\geq \int^{b_1}_{a_1}[(u(t)+u(\sigma(t)))u^{\Delta}(t)w(t)
+\lambda_1(t)w^2(t)u^2(\sigma(t))+kp(t)u^2(\sigma(t))]\Delta t\\
&=\int^{b_1}_{a_1}\Big[(\lambda_1(t))^{1/2}u(\sigma(t))w(t)
+\frac{u(t)+u(\sigma(t))}{2(\lambda_1(t))^{1/2}u(\sigma(t))}u^{\Delta}(t)\Big]^2
\Delta t\\
&\quad +\int^{b_1}_{a_1}\Big[kp(t)u^2(\sigma(t))-\frac{(u(t)
+u(\sigma(t))^2}{4\lambda_1(t)u^2(\sigma(t))}(u^{\Delta}(t))^2\Big]\Delta
t\\
&>\int^{b_1}_{a_1}\Big[kp(t)u^2(\sigma(t))-\frac{(u(t)
+u(\sigma(t))^2}{4\lambda_1(t)u^2(\sigma(t))}(u^{\Delta}(t))^2\Big]\Delta
t,
\end{aligned} \label{e3.7}
\end{equation}
which contradicts \eqref{e3.2}.

In the case of $x(t)<0$ for $t\geq t_0\geq a$, we use the function
$y=-x$ as a positive solution of the dynamic equation $x^{\Delta
\Delta}(t)+p(t)f(x^{\sigma}(t))=-e(t)$ and repeat the above
procedure on the interval $[a_2, b_2]$. This completes the proof of
theorem \ref{th1}.
\end{proof}

\begin{theorem}\label{th2}
Let $xf(x)>0$ for $x\neq 0$ and $|f(x)|\geq |x|^{r}$ for $r>1$.
Assume, in addition, that for any $T\geq a$, there exist constants
$a_1, b_1, a_2, b_2\in \mathbb{T}$ such that \eqref{e3.1} holds. If there
exists $u(t)\in D(a_i, b_i)$ such that
\begin{equation}
\begin{aligned}
&\int^{b_i}_{a_i}\Big\{r(r-1)^{\frac{1-r}{r}}p^{1/r}(t)
|e(t)|^{\frac{r-1}{r}}u^2(\sigma(t)) \\
&-\frac{\mu(t)+t-a_i}{4(t-a_i)}
\big[\frac{u(t)+u(\sigma(t))}{u(\sigma(t))}u^{\Delta}(t)\big]^2\Big\}\Delta
t\geq 0,
\end{aligned} \label{e3.8}
\end{equation}
for $i=1, 2$, then \eqref{e1.4} is oscillatory.
\end{theorem}

\begin{proof}
 As before, we suppose $x(t)>0$, $t\geq t_0\geq a$, be a
nonoscillatory solution of \eqref{e1.4}. Let
$A=p^{1/r}(t)x^{\sigma}(t),
B=\big(\frac{-e(t)}{r-1}\big)^{1/r}$. By the assumption,
we can choose $a_1, b_1\in \mathbb{T}$ such that
$b_1>a_1\geq t_0\geq a$ and $p(t)\geq 0$, $e(t)\leq 0$ for
$t\in [a_1, b_1]$.
Hence, $A>0$, $B>0$ for $r>1$. From lemma \ref{lem1}, we obtain
\begin{equation}
\begin{aligned}
p(t)x^{r}(\sigma(t))-e(t)
&\geq r(r-1)^{\frac{1-r}{r}}p^{1/r}(t)
|e(t)|^{\frac{r-1}{r}}x(\sigma(t)) \\
&=\lambda_2 p^{1/r}(t)|e(t)|^{\frac{r-1}{r}}x(\sigma(t)),
\end{aligned} \label{e3.9}
\end{equation}
where $\lambda_2=r(r-1)^{\frac{1-r}{r}}$ is a constant. By \eqref{e1.4}
and \eqref{e3.9}, we obtain
\begin{equation}
x^{\Delta
\Delta}(t)+\lambda_2p^{1/r}(t)|e(t)|^{\frac{r-1}{r}}x(\sigma(t))
\leq 0. \label{e3.10}
\end{equation}
Let $w(t)=x^{\Delta}(t)/ x(t)$ and use \eqref{e2.1}, \eqref{e2.3} and
\eqref{e3.4}, then
\begin{equation}
w^{\Delta}(t)=\frac{x^{\Delta
\Delta}(t)}{x^{\sigma}(t)}-\frac{x(t)}{x^{\sigma}(t)}
[\frac{x^{\Delta}(t)}{x(t)}]^2 \leq -\lambda_2
p^{1/r}(t)|e(t)|^{\frac{r-1}{r}}-\lambda_1(t)w^2(t),
\label{e3.11}
\end{equation}
where $\lambda_1(t)=\frac{t-a_1}{\mu(t)+t-a_1}$. Let $u(t)\in D(a_1,
b_1)$, product both sides of \eqref{e3.11} by $u^2(\sigma(t))$ and
integrate it from $a_1$ to $b_1$, we get
$$
\int^{b_1}_{a_1}u^2(\sigma(t))w^{\Delta}(t)\Delta t \leq
\int^{b_1}_{a_1}\big[-\lambda_1(t)w^2(t)u^2(\sigma(t)) -\lambda_2
p^{1/r}(t)|e(t)|^{\frac{r-1}{r}}u^2(\sigma(t))\big]\Delta t.
$$
Using integration by parts formula \eqref{e2.4}, and
$u(a_1)=u(b_1)=0$, we have
\begin{align*}
0&=w(t)u^2(t)|^{b_1}_{a_1} \\
&\leq \int^{b_1}_{a_1}\Big[(u(t)+u(\sigma(t)))u^{\Delta}(t)w(t)
 -\lambda_1(t)w^2(t)u^2(\sigma(t))\\
&\quad -\lambda_2 p^{1/r}(t)|e(t)|^{\frac{r-1}{r}}u^2(\sigma(t))\Big]
\Delta t\\
&=-\int^{b_1}_{a_1}\Big[(\lambda_1(t))^{1/2}u(\sigma(t))w(t)
-\frac{u(t)+u(\sigma(t))}{2(\lambda_1(t))^{1/2}u(\sigma(t))}u^{\Delta}(t)
\Big]^2\Delta t\\
&\quad +\int^{b_1}_{a_1}\Big[\frac{(u(t)
+u(\sigma(t)))^2}{4\lambda_1(t)u^2(\sigma(t))}(u^{\Delta}(t))^2
-\lambda_2 p^{1/r}(t)|e(t)|^{\frac{r-1}{r}}u^2(\sigma(t))\Big]\Delta
t\\
&<\int^{b_1}_{a_1}\Big[\frac{(u(t)
+u(\sigma(t)))^2}{4\lambda_1(t)u^2(\sigma(t))}(u^{\Delta}(t))^2
-\lambda_2p^{1/r}(t)|e(t)|^{\frac{r-1}{r}}u^2(\sigma(t))\Big]
\Delta t,
\end{align*}
which contradicts \eqref{e3.8}.
\end{proof}

\begin{theorem}\label{th3}
Let $xf(x)>0$ for $x\neq 0$ and $|f(x)|\geq k|x|^{r}$. Suppose,
furthermore, that for any $T\geq a$, there exist constants $a_1,
b_1, a_2, b_2\in \mathbb{T}$ such that \eqref{e3.1} holds.
If $\mu(t)\leq k't$ and there exists $u(t)\in D(a_i, b_i)$ such that
$$
\int^{b_i}_{a_i}\Big[kp(t)u^2(\sigma(t))
-\frac{1}{4M}\big(\frac{u(t)+u(\sigma(t))}{u(\sigma(t))}u^{\Delta}(t)\big)^2
\Big]\Delta t\geq 0, %\label{e3.12}
$$
for $i=1, 2$ and $M, k, k'$ are some positive constants, then
\begin{enumerate}
\item every unbounded solution of \eqref{e1.4} with $r>1$ is oscillatory.
\item every bounded solution of \eqref{e1.4} with $0<r<1$ is oscillatory.
\end{enumerate}
\end{theorem}

\begin{proof}  As before, we assume $x(t)>0$, $x(\sigma(t))>0$,
$t\geq t_0\geq a$, be a nonoscillatory solution of \eqref{e1.4}. Let
$w(t)=-\frac{x^{\Delta}(t)}{x^{r}(t)}$ for $t\geq t_0$. It follows
from \eqref{e1.4}, the condition $f(x)\geq kx^{r}(t)$ and \eqref{e2.6} that
$w(t)$ satisfies
\begin{equation}
\begin{aligned}
w^{\Delta}(t) &=-\frac{x^{\Delta \Delta}(t)}{x^{r}(\sigma(t))}
+\frac{(x^{\Delta}(t))^2}{x^{r}(t)x^{r}(\sigma(t))}
r\int^{1}_{0}[hx(\sigma(t))+(1-h)x(t)]^{r-1}dh\\
&=p(t)\frac{f(x(\sigma(t)))}{x^{r}(\sigma(t))}-\frac{e(t)}{x^{r}(\sigma(t))}\\
&\quad +r\frac{(x^{\Delta}(t))^2}{x^{r}(t)x^{r}(\sigma(t))}
\int^{1}_{0}[hx(\sigma(t))+(1-h)x(t)]^{r-1}dh.
\end{aligned}
\label{e3.13}
\end{equation}
By the assumption, we can choose $a_1, b_1\in \mathbb{T}$ such that
$b_1>a_1\geq t_0\geq a$ and $p(t)\geq 0, e(t)\leq 0$ for $t\in [a_1,
b_1]$. Then $x^{\Delta \Delta}(t)=e(t)-p(t)f(x^{\sigma}(t))\leq 0$
for $t\in [a_1, b_1]$, and \eqref{e3.13} satisfies
\begin{equation}
w^{\Delta}(t)\geq
kp(t)+r\frac{(x^{\Delta}(t))^2}{x^{r}(t)x^{r}(\sigma(t))}
\int^{1}_{0}[hx(\sigma(t))+(1-h)x(t)]^{r-1}dh.
\label{e3.14}
\end{equation}
There are three cases to be considered

\noindent (i) $x^{\Delta}(t)\geq 0$, $t\in [a_1, b_1]$. Then we obtain
\begin{equation}
\int^{1}_{0}[hx(\sigma(t))+(1-h)x(t)]^{r-1}dh \geq
\int^{1}_{0}x^{r-1}(t)dh =x^{r-1}(t).
\label{e3.15}
\end{equation}
Using \eqref{e3.15} and \eqref{e3.4}, \eqref{e3.14} yields
\begin{equation}
\begin{aligned}
w^{\Delta}(t)
&\geq kp(t)+rx^{r-1}(t)\big[\frac{x(t)}{x(\sigma(t))}\big]^{r}w^2(t)\\
&\geq kp(t)+r\big[\frac{t-a_1}{\mu(t)+t-a_1}\big]^{r}x^{r-1}(t)w^2(t)\\
&= kp(t)+r\lambda^{r}_1(t)x^{r-1}(t)w^2(t), \quad t\in [a_1, b_1],
\end{aligned} \label{e3.16}
\end{equation}
where $\lambda_1(t)=\frac{t-a_1}{\mu(t)+t-a_1}$.

\noindent (ii) $x^{\Delta}(t)<0$, $t\in [a_1, b_1]$. Then we get
\begin{equation}
\int^{1}_{0}[hx(\sigma(t))+(1-h)x(t)]^{r-1}dh \geq
\int^{1}_{0}x^{r-1}(\sigma(t))dh=x^{r-1}(\sigma(t)).
\label{e3.17}
\end{equation}
Using \eqref{e3.17} and \eqref{e3.4}, \eqref{e3.14} yields
\begin{equation}
\begin{aligned}
w^{\Delta}(t)
&\geq kp(t)+rx^{r-1}(t)\frac{x(t)}{x(\sigma(t))}w^2(t) \\
&\geq kp(t)+r\frac{t-a_1}{\mu(t)+t-a_1}x^{r-1}(t)w^2(t)\\
&= kp(t)+r\lambda_1(t)x^{r-1}(t)w^2(t), \quad t\in [a_1, b_1].
\end{aligned}
\label{e3.18}
\end{equation}

\noindent (iii) there exist $a_1<c_1<b_1$ ($c_1\in\mathbb{T}$) such
that $x^{\Delta}(t)\geq 0$, $t\in [a_1, c_1]$ and $x^{\Delta}(t)<0,
t\in (c_1, b_1]$. Proceeding as in the proof of (i)
 and (ii), we obtain that
\begin{gather}
w^{\Delta}(t)\geq kp(t)+r\lambda^{r}_1(t)x^{r-1}(t)w^2(t), \quad
t\in [a_1, c_1], \label{e3.19}
\\
w^{\Delta}(t)\geq kp(t)+r\lambda_1(t)x^{r-1}(t)w^2(t), \quad t\in
(c_1, b_1]. \label{e3.20}
\end{gather}

Next, we consider the following two cases

\noindent (I) If $x$ is an unbounded nonoscillatory solution of \eqref{e1.4}
with $r>1$. Since $\mu(t)\leq k't$ for $k'>0$ is a positive
constant, then there exists a positive constant
$0<k''<\frac{1}{k'+1}$ such that
$k''<\frac{t-a_1}{\mu(t)+t-a_1}\leq 1$ for $t$ large enough,
from \eqref{e3.16}, \eqref{e3.18}, \eqref{e3.19}, and \eqref{e3.20}, we get
\begin{equation}
w^{\Delta}(t)\geq kp(t)+r\lambda^{r}_1(t)x^{r-1}(t)w^2(t).
\label{e3.21}
\end{equation}
Since $x^{\Delta\Delta}(t)\leq 0, t\in [a_1, b_1]$, then there
exists a constant $M_1>0$ such that $x(t)\geq M_1$ on $[a_1, b_1]$,
such that
\begin{equation}
r\lambda^{r}_1(t)x^{r-1}(t)\geq r\lambda^{r}_1(t)M^{r-1}_1\geq M,
\quad t\in [a_1, b_1],
\label{e3.22}
\end{equation}
where $M>0$ is a constant. Using \eqref{e3.21} and \eqref{e3.22}, and proceeding
as in the proof of theorem \ref{th1}, we obtain the desired
contradiction.

\noindent (II) If $x$ is a bounded nonoscillatory solution of
\eqref{e1.4} with $0<r<1$ on $[a_1, b_1]$. Since
$0<k''<\frac{t-a_1}{\mu(t)+t-a_1}\leq 1$ for $t$ large enough, from
\eqref{e3.16}, \eqref{e3.18}, \eqref{e3.19}, and \eqref{e3.20}, we
obtain
$$
w^{\Delta}(t)\geq kp(t)+r\lambda_1(t)x^{r-1}(t)w^2(t).
%\label{e3.23}}
$$
Since $x^{\Delta\Delta}(t)\leq 0, t\in [a_1, b_1]$, then there
exists a constant $M_2>0$ such that $x(t)\leq M_2$ on $[a_1, b_1]$,
hence
$$
r\lambda_1(t)x^{r-1}(t)\geq r\lambda_1(t)M^{r-1}_2\geq M', \quad
t\in [a_1, b_1],
%\label{e3.24}
$$
where $M'>0$ is a constant. The rest of the proof is similar to
that in the previous case and we obtain the desired contradiction.
\end{proof}

\section{Example}

Since the time scale $\mathbb{P}_{a,b}=\cup^{\infty}_{n=0}[n(a+b), n(a+b)+a]$
can be used to study many models of real world, for instance,
population in biological communities, electric circuit and so on,
we give an example in such a time scale to demonstrate how the
theory may be applied to specific problems.

  Consider the  forced second order dynamic equation
\begin{equation}
x^{\Delta\Delta}(t)+m \\sin t x(\sigma(t))=\cos t, \quad \text{for }
 t\in \mathbb{P}_{\pi, \pi}=\cup^{\infty}_{n=0}[2n\pi, (2n+1)\pi],
\label{e4.1}
\end{equation}
with the transition condition
\begin{equation}
x(2n\pi)=x((2n-1)\pi), \quad n\geq 1, \label{e4.2}
\end{equation}
where $m>0$ is a constant, $p(t)=m \sin t$, $e(t)=\cos t$,
$f(x(\sigma(t)))=x(\sigma(t))$. For any $T\geq 0$, if we choose
$a_1=2n\pi+\frac{\pi}{2}$, $b_1=2n\pi+\frac{3\pi}{4}$,
$a_2=2n\pi+\frac{\pi}{4}$, $b_2=2n\pi+\frac{\pi}{2}$,
($a_i, b_i \in \mathbb{P}_{\pi, \pi}$, $i=1, 2$)
 such that $a_i\geq T$ for
sufficiently large $n, i=1, 2$, then we have $p(t)\geq 0$ for
$t\in [a_1, b_1]\bigcup [a_2, b_2]$, $e(t)\leq 0$ for $t\in [a_1, b_1]$, and
$e(t)\geq 0$ for $t\in [a_2, b_2]$. Choose $u(t)=sin2t cos2t$, then
$u(t)\in D(a_i, b_i), i=1, 2$. Furthermore, we have $\sigma(t)=t$,
$\mu(t)=0$ for $t\in [a_i, b_i], i=1, 2$. Noting that for $i=1, 2$,
\begin{align*}
&\int^{b_i}_{a_i}\big\{kp(t)u^2(\sigma(t))
-\frac{\mu(t)+t-a_i}{4(t-a_i)}
\big[\frac{u(t)+u(\sigma(t))}{u(\sigma(t))}u^{\Delta}(t)\big]^2\big\}\Delta
t\\
&=\int^{b_i}_{a_i}\big[p(t)u^2(t)-(u'(t))^2\big]dt\\
&=\int^{b_i}_{a_i} m\sin t \sin^{2}(2t)
\cos^{2}(2t)-4\cos^{2}(4t)\big]dt.
\end{align*} %\label{e4.3}
On the other hand, we have
$$
\int^{b_i}_{a_i}4\cos^{2}(4t)dt=\frac{\pi}{2},
$$
and
$$
\int^{b_i}_{a_i}m\sin t \sin^{2}(2t) \cos^{2}(2t) dt
=\frac{\sqrt{2}}{2} m\big[\frac{1}{8}-\frac{1}{9\times
16}+\frac{1}{7\times 16}\big].
$$
Then, $\int^{b_i}_{a_i}m\sin t \sin^{2}(2t) \cos^{2}(2t) dt\geq
\pi/ 2$ for sufficiently large $m$, hence \eqref{e3.2} holds. By
Theorems \ref{th1}, we obtain that \eqref{e4.1} and \eqref{e4.2} is oscillatory.
However, the results in  Saker \cite{Saker2} and Bohner and
Saker \cite{Bohner2} cannot be applied the oscillation of \eqref{e4.1}
and \eqref{e4.2}.

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\end{document}